Question

Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a sample of size 322 with 270 successes. Enter your answer using decimals (not percents) accurate to three decimal places.

Answer 1

Answer with explanation:

Sample Size=322

Success=270

Probability of Success(p)

`=(270)/(322)\\\\=(135)/(161)`

At 98% confidence interval , population proportion will be

`=z_{98 \text{Percent}}* \sqrt{(p(1-p))/(n)}\\\\=0.8365 *\sqrt{((270)/(322)(1-(270)/(322)))/(322)}\\\\=0.8365 *\sqrt{((270)/(322)*((52)/(322)))/(322)}\\\\=(0.8365)/(322) * √(270 * 52)\\\\=0.00259782 * √(14040)\\\\=0.00259782 * 118.490\\\\=0.307817`

= 0.308( approx to three decimal places.)